Question

Two identical chargedloop of radius r having charge q and minus q are placed coaxially at 2r distance apart find the electric field intensity at the centre of each, at the middle point of the line joining the center of the loops, at distance r along the axis but outside position

Answer 1

Explanation:

Let the charge on A and B be = Q each.

Let the distance of separation between them be = 2R

According to Coloumb’s law of electrostats,

Force between them F1 = KQ2/(2R)2

F1 = KQ2/4R2 this force is given in the problem as 2*10-5N.

When an uncharged sphere C is brought in contact with A first, the charges get equally shared between them , hence each sphere acquires a charge Q/2.

Now force between A &C after it is placed at mid point ie at distance R from A is:

F2 = K(Q/2 )*( Q/2)/R2

F2 = KQ2/4R2

Force between C and B will be :

F3 = (K(Q/2) * Q)/(R)2

F3 = KQ2/2R2

Clearly, F3>F2 therefore F3-F2:-

Hence KQ2/R2( 1/2 - 1/4 )

= KQ2/R2( ¼ )

= KQ2/4R2

But this is the given force F1 from eqn 1 hence the net force on C = 2*10-5N

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Answer 2

❇️❇️❇️❇️❇️❇️❇️❇️❇️❇️❇️❇️❇️❇️❇️

✒️Your Question :-

➡️ Two identical chargedloop of radius r having charge q and minus q are placed coaxially at 2r distance apart find the electric field intensity at the centre of each, at the middle point of the line joining the center of the loops, at distance r along the axis but outside position

____________________________

✒️Answer :-

According to Coloumb’s law of electrostats,

Force between them F1 = KQ2/(2R)2

F1 = KQ2/4R2 this force is given in the problem as 2*10-5N.

When an uncharged sphere C is brought in contact with A first, the charges get equally shared between them , hence each sphere acquires a charge Q/2.

Now force between A &C after it is placed at mid point ie at distance R from A is:

F2 = K(Q/2 )*( Q/2)/R2

F2 = KQ2/4R2

Force between C and B will be :

F3 = (K(Q/2) * Q)/(R)2

F3 = KQ2/2R2

Clearly, F3>F2 therefore F3-F2:-

Hence KQ2/R2( 1/2 - 1/4 )

= KQ2/R2( ¼ )

= KQ2/4R2

But this is the given force F1 from eqn 1 hence the net force on C = 2*10-5N

___________________________

➡️ hope this helps you ❗️❗️

✔️✔️

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